, 2013

Solution:

Since $AD$ is an angle bisector, $D$ is the midpoint of $\operatorname{arc} BC$ opposite $A$ on $\omega$. It is well-known that $B, I$, and $C$ lie on a circle centered at $D$. Thus $BD = DC = DI$. Applying Ptolemy's theorem to cyclic quadrilateral $ABDC$, we get
$$AB \cdot DC + AC \cdot BD = AD \cdot BC = AD \cdot (AB + AC)/2$$
Using $BD = DC$ we have immediately that $AD = 2 BD = 2 DI$ so $I$ is the midpoint of $AD$ as desired.

Solution:

Let $P$ and $Q$ be the midpoints of $AB$ and $AC$, and take the point $E$ on segment $BC$ such that $BE = BP$. Note that $CE = AB - BE = AB - BP = \frac{AB + AC}{2} - \frac{AB}{2} = \frac{AC}{2} = CQ$, so triangles $BPE$ and $CQE$ are isosceles. In addition, $\frac{BE}{EC} = \frac{AB/2}{AC/2} = \frac{AB}{AC}$, so by the angle bisector theorem, $AE$ bisects $\angle CAB$, whence $E$ must lie on the bisector of $\angle A$.

Since triangles $BPE$ and $CQE$ are isosceles, the bisectors of angles $B$ and $C$ are the perpendicular bisectors of segments $PE$ and $EQ$, respectively. Thus, the circumcenter of $\triangle PQE$ is $I$, so the perpendicular bisector of $PQ$ meets the bisector of $\angle A$ at $I$.

Furthermore, since $\angle DAB = \angle CAD$, arcs $\widehat{BD}$ and $\widehat{DC}$ have the same measure, so $BD = DC$, whence the perpendicular bisector of $BC$ meets the bisector of $\angle A$ at $D$. A homothety centered at $A$ with factor $1/2$ maps $BC$ to $PQ$, and so maps $D$ to $I$. Thus, $D$ is the midpoint of $AI$.

Solution:

Let $a = BC$, $b = CA$, and $c = AB$, let $r$ and $R$ denote the lengths of the inradius and circumradius of $\triangle ABC$, respectively, let $E$ be the intersection of segments $AD$ and $BC$, and let $O$ be the circumcenter of $\triangle ABC$. $I$ is the midpoint of chord $AD$ if and only if $OI \perp AD$, which is true if and only if $OA^2 = OI^2 + AI^2$. By Euler's distance formula, $OI^2 = R(R - 2r)$, and by Stewart's theorem and the angle bisector theorem we can find
$$AI^2 = \left(\frac{b + c}{a + b + c}\right)^2 bc \left(1 - \left(\frac{a}{b + c}\right)^2\right) = \frac{bc}{3}.$$
Thus, it remains to show that $6Rr = bc$.

Now, we use the well-known formulas $\frac{abc}{4K} = R$ and $K = rs$ to get $abc = 4Rrs$, where $K$ is the area of $\triangle ABC$ and $s$ is its semiperimeter. We have
$$bc = Rr \frac{4s}{a} = Rr \frac{2a + 2(b + c)}{a} = Rr \frac{2a + 2(2a)}{a} = 6Rr,$$
as desired.